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Aleonysh [2.5K]
2 years ago
8

A consumer group wants to know if an automobile insurance company with thousands of customers has an average insurance payout fo

r all their customers that is greater than $500 per insurance claim. They know that most customers have zero payouts and a few have substantial payouts. The consumer group collects a random sample of 18 customers and computes a mean payout per claim of $579.80 with a standard deviation of $751.30.
Is it appropriate for the consumer group to perform a hypothesis test for the mean payout of all customers?
A. Yes, it is appropriate because the population standard deviation is unknown.
B. Yes, it is appropriate because the sample size is large enough, so the condition that the sampling distribution of the sample mean be approximately normal is satisfied.
C. No, it is not appropriate because the sample is more than 10 percent of the population, so a condition for independence is not satisfied.
D. No, it is not appropriate because the standard deviation is greater than the mean payout, so the condition that the sampling distribution of the sample mean be approximately normal is not satisfied.
E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.
Mathematics
1 answer:
zhenek [66]2 years ago
7 0

Answer:

E. No, it is not appropriate because the distribution of the population is skewed and the sample size is not large enough to satisfy the condition that the sampling distribution of the sample mean be approximately normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question:

Standard deviation larger than the sample mean means that the distribution is skewed.

By the Central Limit Theorem, when the distribution is skewed, normality is assumed for samples sizes of 30 or higher. In this question, the sample is of 18, which is less than 30, so the hypothesis test is not appropriated, and the correct answer is given by option E.

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235.5‬

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A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far
ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3} (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2}) (1b)

Where:

x, y - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

r_{1}, r_{2} - Length of each vector, in kilometers.

\theta_{1}, \theta_{2} - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that r_{1} = 42\,km, r_{2} = 28\,km, \theta_{1} = 32^{\circ} and \theta_{2} = 154^{\circ}, then the resulting coordinates of the final position of the surveyor is:

(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})

(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]

(x,y) = (10.452, 34.531)\,[km]

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}

\theta \approx 16.840^{\circ}

And the distance from the camp is calculated by the Pythagorean Theorem:

r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}

r = 36.078\,km

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

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3 years ago
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